```
public int findMaxConsecutiveOnes(int[] nums) {
if (nums == null || nums.length == 0) return 0;
int maxCount = 0, runningCount = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] == 0) {
maxCount = Math.max(maxCount, runningCount);
runningCount = 0; // reset running count
} else {
runningCount++; // increment running count
}
}
maxCount = Math.max(maxCount, runningCount); // in case nums ends with a trail of 1s
return maxCount;
}
```

This solution beats other solutions like this one simply because it inverts where it calculates the maxCount (presumably the OJ has input test cases with a large number of 1s that reduces the number of times `Math.max(..)`

has to be called in my code.